| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2008 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Determine constant from stationary point condition |
| Difficulty | Moderate -0.3 This is a straightforward multi-part stationary points question requiring routine differentiation, substitution to find k, second derivative test, and solving a quadratic. All steps are standard C1 techniques with no problem-solving insight needed, making it slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = 3x^2 - 2kx + 1\) | B1 | One term correct |
| B1 | Fully correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(3x^2 - 2kx + 1 = 0\) when \(x = 1\) | M1 | Their \(\frac{dy}{dx} = 0\) set equal to zero |
| \(3 - 2k + 1 = 0\) | M1 | \(x = 1\) substituted into their \(\frac{dy}{dx} = 0\) |
| \(k = 2\) | A1\(\checkmark\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{d^2y}{dx^2} = 6x - 4\) | M1 | Substitutes \(x = 1\) into their \(\frac{d^2y}{dx^2}\) and looks at sign |
| When \(x = 1\), \(\frac{d^2y}{dx^2} > 0\) \(\therefore\) minimum point | A1 | States minimum CWO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(3x^2 - 4x + 1 = 0\) | M1 | Their \(\frac{dy}{dx} = 0\) |
| \((3x-1)(x-1) = 0\) | M1 | Correct method to solve 3-term quadratic |
| \(x = \frac{1}{3},\ x = 1\) | ||
| \(x = \frac{1}{3}\) | A1 | WWW at any stage |
## Question 8(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = 3x^2 - 2kx + 1$ | B1 | One term correct |
| | B1 | Fully correct |
---
## Question 8(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3x^2 - 2kx + 1 = 0$ when $x = 1$ | M1 | Their $\frac{dy}{dx} = 0$ set equal to zero |
| $3 - 2k + 1 = 0$ | M1 | $x = 1$ substituted into their $\frac{dy}{dx} = 0$ |
| $k = 2$ | A1$\checkmark$ | |
---
## Question 8(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{d^2y}{dx^2} = 6x - 4$ | M1 | Substitutes $x = 1$ into their $\frac{d^2y}{dx^2}$ and looks at sign |
| When $x = 1$, $\frac{d^2y}{dx^2} > 0$ $\therefore$ minimum point | A1 | States minimum **CWO** |
---
## Question 8(iv):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3x^2 - 4x + 1 = 0$ | M1 | Their $\frac{dy}{dx} = 0$ |
| $(3x-1)(x-1) = 0$ | M1 | Correct method to solve 3-term quadratic |
| $x = \frac{1}{3},\ x = 1$ | | |
| $x = \frac{1}{3}$ | A1 | **WWW** at any stage |
---8 The curve $y = x ^ { 3 } - k x ^ { 2 } + x - 3$ has two stationary points.\\
(i) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.\\
(ii) Given that there is a stationary point when $x = 1$, find the value of $k$.\\
(iii) Determine whether this stationary point is a minimum or maximum point.\\
(iv) Find the $x$-coordinate of the other stationary point.
\hfill \mbox{\textit{OCR C1 2008 Q8 [10]}}